Asked by peter
a cylinder of diameter 1cm at 30 degrees celcius is to be slid into a steel plate. the hole has a diameter 0.9997cm at 30 degrees celcius. to what temperature must the plate be heated? for steel
Answers
Answered by
Steve
As I recall, we have the thermal expansion equation is
dL = Lα dT
where α for steel is 13.0*10^-6 m/m /°K
So, since we want the diameter of the hole to increase by 0.0003 cm, we want the circumference to increase by .0003pi = .0009425 cm = 9.425*10^-6 m
So, now we just have to plug those numbers into our equation:
9.425*10^-6 = (3.14065*10^-2)(13.0*10^-6) dT
dT = 23.08 °K
So, we have to raise the temperature by 23°, to 53°C
dL = Lα dT
where α for steel is 13.0*10^-6 m/m /°K
So, since we want the diameter of the hole to increase by 0.0003 cm, we want the circumference to increase by .0003pi = .0009425 cm = 9.425*10^-6 m
So, now we just have to plug those numbers into our equation:
9.425*10^-6 = (3.14065*10^-2)(13.0*10^-6) dT
dT = 23.08 °K
So, we have to raise the temperature by 23°, to 53°C
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